An Armature-Controlled DC Motor Consider the dynamic system shown in Figure 6.66, which represents an armature-controlled DC motor. Assume that the armature inductance is negligibly small, that is, La = 0. The system dynamics can be expressed as Raia + Keθm = va, Imθm + Bmθm - Ktia = 0, where the

Question

An Armature-Controlled DC Motor

Consider the dynamic system shown in Figure 6.66, which represents an armature-controlled DC motor. Assume that the armature inductance is negligibly small, that is, [latex]L_{a}=0[/latex]. The system dynamics can be expressed as

[latex] \begin{gathered} R_{\mathrm{a}} i_{\mathrm{a}}+K_{\mathrm{e}} \dot{ heta}_{\mathrm{m}}=v_{\mathrm{a}^{\prime}} \ I_{\mathrm{m}} \ddot{ heta}_{\mathrm{m}}+B_{\mathrm{m}} \dot{ heta}_{\mathrm{m}}-K_{\mathrm{t}} i_{\mathrm{a}}=0 \end{gathered} [/latex]

where the armature resistance is [latex]R_{\mathrm{a}}=0.5 \Omega[/latex], the back emf constant is [latex]K_{\mathrm{e}}=0.05 \mathrm{~V} \cdot \mathrm{s} / \mathrm{rad}[/latex], the torque constant is [latex]K_{\mathrm{t}}=0.05 \mathrm{~N} \cdot \mathrm{m} / \mathrm{A}[/latex], the mass moment of inertia of the motor is [latex]I_{\mathrm{m}}=0.00025 \mathrm{~kg} \cdot \mathrm{m}^{2}[/latex], the coefficient of the torsional viscous damping of the motor is [latex]B_{\mathrm{m}}=0.0001 \mathrm{~N} \cdot \mathrm{m} \cdot \mathrm{s} / \mathrm{rad}[/latex], and the applied voltage is [latex]v_{\mathrm{a}}=10 \mathrm{~V}[/latex].

a. Denote [latex]\omega_{\mathrm{m}}=\dot{ heta}_{\mathrm{m}}[/latex]. Following Figure 6.45 , build a Simulink block diagram using the given equations and find the armature current output [latex]i_{\mathrm{a}}(t)[/latex] and the rotor speed [latex]\omega_{\mathrm{m}}(t)[/latex].

b. Assume zero initial conditions, determine the transfer functions [latex]I_{\mathrm{a}}(s) / V_{\mathrm{a}}(s)[/latex] and [latex]\Omega_{\mathrm{m}}(s) / V_{\mathrm{a}}(s)[/latex]. Build a Simulink block diagram using these two transfer functions and find the armature current output [latex]i_{\mathrm{a}}(t)[/latex] and the rotor speed [latex]\omega_{\mathrm{m}}(t)[/latex].

c. Choose [latex] heta_{\mathrm{m}}[/latex] and [latex]\dot{ heta}_{\mathrm{m}}[/latex] as state variables and determine the state-space form of the system. Build a Simulink block diagram based on the state-space form and find the armature current output [latex]i_{\mathrm{a}}(t)[/latex] and the rotor speed [latex]\omega_{\mathrm{m}}(t)[/latex].

d. Build a Simscape model of the DC motor and find the armature current output [latex]i_{a}(t)[/latex] and the rotor speed [latex]\omega_{\mathrm{m}}(t)[/latex].

Accepted Answer

a. This electromechanical system includes an armature circuit

[latex]R_{\mathrm{a}} i_{\mathrm{a}}=v_{\mathrm{a}}-\mathrm{e}_{\mathrm{b}}[/latex]

a rotational system

[latex]I_{\mathrm{m}} \dot{\omega}_{\mathrm{m}}+B_{\mathrm{m}} \omega_{\mathrm{m}}= au_{\mathrm{m}}[/latex]

and couplings between the electrical and mechanical subsystems

[latex]\mathrm{e}_{\mathrm{b}}=K_{\mathrm{e}} \omega_{\mathrm{m}} \quad au_{\mathrm{m}}=K_{\mathrm{t}} i_{\mathrm{a}}[/latex]

Following Figure 6.45, we can construct a Simulink block diagram (see Figure 6.67), which shows the major components of the DC motor and their interconnections. The dynamics of the mechanical rotational system is represented using a Transfer Fen block. The armature resistance, torque constant, and back emf constant are represented using Gain blocks.

b. Taking Laplace transform of the two equations given and denoting [latex]s \Theta_{\mathrm{m}}(s)=\Omega_{\mathrm{m}}(s)[/latex] provides

[latex] \begin{gathered} R_{\mathrm{a}} I_{\mathrm{a}}(s)+K_{\mathrm{e}} \Omega(s)=V_{\mathrm{a}}(s), \ I_{\mathrm{m}} s \Omega_{\mathrm{m}}(s)+B_{\mathrm{m}} \Omega_{\mathrm{m}}(s)-K_{\mathrm{t}} I_{\mathrm{a}}(s)=0 . \end{gathered} [/latex]

Using Cramer's rule to solve for the transfer functions [latex]I_{\mathrm{a}}(s) / V_{\mathrm{a}}(s)[/latex] and [latex]\Omega_{\mathrm{m}}(s) / V_{\mathrm{a}}(s)[/latex] yields

[latex] \begin{gathered} \frac{I_{\mathrm{a}}(s)}{V_{\mathrm{a}}(s)}=\frac{I_{\mathrm{m}} s+B_{\mathrm{m}}}{R_{\mathrm{a}} I_{\mathrm{m}} s+R_{\mathrm{a}} B_{\mathrm{m}}+K_{\mathrm{t}} K_{\mathrm{e}}}, \ \frac{\Omega_{\mathrm{m}}(s)}{V_{\mathrm{a}}(s)}=\frac{K_{\mathrm{t}}}{R_{\mathrm{a}} I_{\mathrm{m}} s+R_{\mathrm{a}} B_{\mathrm{m}}+K_{\mathrm{t}} K_{\mathrm{e}}}, \end{gathered} [/latex]

both of which can easily be represented using a Transfer Fcn block (see Figure 6.68).

c. Let [latex]x_{1}= heta_{\mathrm{m}}, x_{2}=\dot{ heta}_{\mathrm{m}^{\prime}}[/latex] and [latex]u=v_{\mathrm{a}}[/latex] and the state-variable equations are

[latex] \begin{gathered} \dot{x}_{1}=\dot{ heta}_{\mathrm{m}}=x_{2^{\prime}} \ \dot{x}_{2}=\ddot{ heta}_{\mathrm{m}}=\frac{K_{\mathrm{t}}}{I_{\mathrm{m}}} i_{\mathrm{a}}-\frac{B_{\mathrm{m}}}{I_{\mathrm{m}}} \dot{ heta}_{\mathrm{m}}=\frac{K_{\mathrm{t}}}{I_{\mathrm{m}}}\left(\frac{v_{\mathrm{a}}-K_{\mathrm{e}} \dot{ heta}_{\mathrm{m}}}{R_{\mathrm{a}}} ight)-\frac{B_{\mathrm{m}}}{I_{\mathrm{m}}} \dot{ heta}_{\mathrm{m}} \ =-\left(\frac{K_{\mathrm{t}} K_{\mathrm{e}}}{I_{\mathrm{m}} R_{\mathrm{a}}}+\frac{B_{\mathrm{m}}}{I_{\mathrm{m}}} ight) x_{2}+\frac{K_{\mathrm{t}}}{I_{\mathrm{m}} R_{\mathrm{a}}} u \end{gathered} [/latex]

or in matrix form

[latex] \left\{\begin{array}{l} \dot{x}_{1} \ \dot{x}_{2} \end{array} ight\}=\left[\begin{array}{cc} 0 & 1 \ 0 & -\left(\frac{K_{\mathrm{t}} K_{\mathrm{e}}}{I_{\mathrm{m}} R_{\mathrm{a}}}+\frac{B_{\mathrm{m}}}{I_{\mathrm{m}}} ight) \end{array} ight]\left\{\begin{array}{l} x_{1} \ x_{2} \end{array} ight\}+\left[\begin{array}{c} 0 \ \frac{K_{\mathrm{t}}}{I_{\mathrm{m}} R_{\mathrm{a}}} \end{array} ight] u . [/latex]

The output equations are

[latex] \begin{gathered} y_{1}=i_{\mathrm{a}}=\frac{v_{\mathrm{a}}-K_{\mathrm{e}} \dot{ heta}_{\mathrm{m}}}{R_{\mathrm{a}}}=-\frac{K_{\mathrm{e}}}{R_{\mathrm{a}}} x_{2}+\frac{1}{R_{\mathrm{a}}} u, \ y_{2}=\dot{ heta}_{\mathrm{m}}=x_{2} \end{gathered} [/latex]

or in matrix form

[latex] \left\{\begin{array}{l} y_{1} \ y_{2} \end{array} ight\}=\left[\begin{array}{cc} 0 & -\frac{K_{\mathrm{e}}}{R_{\mathrm{a}}} \ 0 & 1 \end{array} ight]\left\{\begin{array}{l} x_{1} \ x_{2} \end{array} ight\}+\left[\begin{array}{c} \frac{1}{R_{\mathrm{a}}} \ 0 \end{array} ight] u [/latex]

The system can be represented using a State-Space block (see Figure 6.69) with A, B, C, D matrices defined above. A Demux block from the Simulink library of Signal Routing is used to split the output vector signal into two scalar signals [latex]i_{\mathrm{a}}(t)[/latex] and [latex]\omega_{\mathrm{m}}(t)[/latex].

d. The Simscape block diagram of the DC motor (see Figure 6.70) consists of elements from two domains, electrical and mechanical rotational. Note that each domain requires at least one reference block. As shown in Figure 6.70, both Electrical Reference and Mechanical Rotational Reference blocks are attached to the appropriate circuit.

Define all the parameters in MATLAB Workspace and run the Simulink or Simscape models. The result can be plotted as shown in Figures 6.71 and 6.72 .

Question 6.17: An Armature-Controlled DC Motor Consider the dynamic system ...

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